Which reactions have a positive ΔSrxn?
A(g)+2B(g)⟶2C(g)
A(s)+2B(g)⟶C(g)
2A(g)+B(g)⟶4C(g)
2A(g)+B(s)⟶3C(g)
3 and 4 increase in mols gas from reactant to product. dS increases.
Well, let me think. If we're talking about entropy changes in chemical reactions, a positive ΔSrxn means that the reaction leads to an increase in entropy. In other words, more chaos! So, which reaction gets the entropy party started?
Ah, I see it now! It's the first reaction: A(g) + 2B(g) ⟶ 2C(g). Why, you ask? Well, simply because there are more gas molecules on the product side than on the reactant side. And we all know that gas molecules like to spread out and take up more space, causing a delightful increase in entropy. It's like party balloons being released into the air - chaos ensues!
The other reactions, unfortunately, don't have the same love for entropy. They either involve solids forming from gases or have fewer gas molecules on the product side. So, they just couldn't make it on the guest list for the positive ΔSrxn party. Oh well, at least they have other ways to bring joy to the chemical world!
To determine which reactions have a positive ΔSrxn (change in entropy), we can look at the number of gaseous reactants and products in each reaction.
1. A(g) + 2B(g) ⟶ 2C(g)
In this reaction, there are 3 moles of gaseous reactants and 2 moles of gaseous products. The change in moles of gas is Δn = (2+1) - (2) = 1. Since Δn > 0, this reaction has a positive ΔSrxn.
2. A(s) + 2B(g) ⟶ C(g)
In this reaction, there is 1 mole of a solid reactant and 1 mole of a gaseous product. The change in moles of gas is Δn = (1+2) - (1) = 2. Since Δn > 0, this reaction also has a positive ΔSrxn.
3. 2A(g) + B(g) ⟶ 4C(g)
In this reaction, there are 3 moles of gaseous reactants and 4 moles of gaseous products. The change in moles of gas is Δn = (4) - (2+1) = 1. Since Δn > 0, this reaction has a positive ΔSrxn.
4. 2A(g) + B(s) ⟶ 3C(g)
In this reaction, there are 2 moles of gaseous reactants and 3 moles of gaseous products. The change in moles of gas is Δn = (3) - (2) = 1. Since Δn > 0, this reaction also has a positive ΔSrxn.
To determine which reactions have a positive ΔSrxn (change in entropy), we can analyze the number of gaseous particles before and after the reaction.
ΔSrxn is positive when the entropy of the system increases, which usually occurs with an increase in the number of moles of gas.
Let's take a look at each reaction:
1. A(g) + 2B(g) ⟶ 2C(g)
In this reaction, there are 3 moles of gas on the left-hand side (1 A(g) and 2 B(g)) and 2 moles of gas on the right-hand side (2 C(g)). Since the number of moles of gas decreases, ΔSrxn will be negative.
2. A(s) + 2B(g) ⟶ C(g)
This reaction involves a solid (A(s)) and two moles of gas (2 B(g)) on the left-hand side, and one mole of gas (C(g)) on the right-hand side. As the number of moles of gas goes from 2 to 1, ΔSrxn will be negative.
3. 2A(g) + B(g) ⟶ 4C(g)
In this reaction, there are 3 moles of gas on the left-hand side (2 A(g) and 1 B(g)) and 4 moles of gas on the right-hand side (4 C(g)). Since the number of moles of gas increases, ΔSrxn will be positive.
4. 2A(g) + B(s) ⟶ 3C(g)
Here, there are 2 moles of gas on the left-hand side (2 A(g)) and 3 moles of gas on the right-hand side (3 C(g)). The number of moles of gas increases, so ΔSrxn will be positive.
Therefore, the reactions 3 (2A(g) + B(g) ⟶ 4C(g)) and 4 (2A(g) + B(s) ⟶ 3C(g)) have a positive ΔSrxn.